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Date	November 2017	Marks available	6	Reference code	17N.1.AHL.TZ0.H_10
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Show that	Question number	H_10	Adapted from	N/A

Question

Chloe and Selena play a game where each have four cards showing capital letters A, B, C and D.
Chloe lays her cards face up on the table in order A, B, C, D as shown in the following diagram.

N17/5/MATHL/HP1/ENG/TZ0/10

Selena shuffles her cards and lays them face down on the table. She then turns them over one by one to see if her card matches with Chloe’s card directly above.
Chloe wins if no matches occur; otherwise Selena wins.

Chloe and Selena repeat their game so that they play a total of 50 times.
Suppose the discrete random variable X represents the number of times Chloe wins.

Show that the probability that Chloe wins the game is $\frac{3}{8}$ .

[6]

a.

Determine the mean of X.

[3]

b.i.

Determine the variance of X.

[2]

b.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

number of possible “deals” $= 4! = 24$ A1

consider ways of achieving “no matches” (Chloe winning):

Selena could deal B, C, D (ie, 3 possibilities)

as her first card R1

for each of these matches, there are only 3 possible combinations for the remaining 3 cards R1

so no. ways achieving no matches $= 3 \times 3 = 9$ M1A1

so probability Chloe wins $= \frac{9}{{23}} = \frac{3}{8}$ A1AG

METHOD 2

number of possible “deals” $= 4! = 24$ A1

consider ways of achieving a match (Selena winning)

Selena card A can match with Chloe card A, giving 6 possibilities for this happening R1

if Selena deals B as her first card, there are only 3 possible combinations for the remaining 3 cards. Similarly for dealing C and dealing D R1

so no. ways achieving one match is $= 6 + 3 + 3 + 3 = 15$ M1A1

so probability Chloe wins $= 1 - \frac{{15}}{{24}} = \frac{3}{8}$ A1AG

METHOD 3

systematic attempt to find number of outcomes where Chloe wins (no matches)

(using tree diag. or otherwise) M1

9 found A1

each has probability $\frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times 1$ M1

$= \frac{1}{{24}}$ A1

their 9 multiplied by their $\frac{1}{{24}}$ M1A1

$= \frac{3}{8}$ AG

[6 marks]

a.

$X \sim {\text{B}}\left( {50,{\text{ }}\frac{3}{8}} \right)$ (M1)

$\mu = np = 50 \times \frac{3}{8} = \frac{{150}}{8}{\text{ }}\left( { = \frac{{75}}{4}} \right){\text{ }}( = 18.75)$ (M1)A1

[3 marks]

b.i.

${\sigma ^2} = np(1 - p) = 50 \times \frac{3}{8} \times \frac{5}{8} = \frac{{750}}{{64}}{\text{ }}\left( { = \frac{{375}}{{32}}} \right){\text{ }}( = 11.7)$ (M1)A1

[2 marks]

b.ii.

Examiners report

[N/A]

a.

[N/A]

b.i.

[N/A]

b.ii.

Syllabus sections

Topic 4—Statistics and probability » SL 4.8—Binomial distribution

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